Suppose each player is randomly and secretly assigned a number between $1$ and $6$ and then a communal dice is rolled four times. After each roll, you can place a bet or fold. After the fourth bet is placed, the person who has the most die facing their number wins.

As each game progresses, the game can move in your favour or against you (e.g., 1/1, 1/2, 2/3, 3/4). If you bet too high at the beginning, the progression of the game could move against you. If you bet too low, it would be difficult to increase the pot at the end because the other players could fold without losing anything. There are infinitely many games available and a finite buy-in amount but an infinite amount of money to replenish the buy-in.

How do you calculate the optimum betting amount?

This relates to one aspect of Texas Hold 'Em poker betting odds, in abstract terms. The other aspects of the betting odds can be ignored.

Any guidelines will be appreciated. I think the question could be simplified to:

Does an initial bet of $10\%$ of capital on $10\%$ odds with a $20\times$ return approach $0$ or $\infty$? What are the limits in this case?

  • $\begingroup$ As with texas holdem, the optimal amount depends on the other players, if you bet too high they might fold and you win nothing. So to calculate the optimum, can you find a way to define that optimum mathematically? Should we assume that each player follows the exact same (optimal) strategy, or do you have different strategies that might be good or bad against one another in mind? $\endgroup$
    – Dirk
    Apr 25, 2019 at 13:57
  • $\begingroup$ That is a very good question. I guess if everyone follows the same strategy the odds would even out over time. So I'm particularly interested in how the betting strategy could give low cards an advantage over high cards. Let's say those who pick high numbers win when there's a draw, as in poker. How would someone with a low number progressively bet to sneak past someone who has a high number without betting too much, e.g. betting $10\%$ of capital when there's a $10\%$ chance of winning. $\endgroup$
    – dataphile
    Apr 25, 2019 at 14:06
  • $\begingroup$ Even though there's a $10\%$ of winning, when that materialises it could result in a 20x winning with enough players and escalated betting as the odds increase to $100\%$ after the final throw. $\endgroup$
    – dataphile
    Apr 25, 2019 at 14:21
  • $\begingroup$ @dataphile I am not clear about the set up here. Can you see my dice rolls ? Can I see your rolls ? $\endgroup$
    – gandalf61
    Apr 25, 2019 at 14:33
  • 1
    $\begingroup$ See en.wikipedia.org/wiki/Kelly_criterion — the strategy that maximizes the expected logarithm of wealth $\endgroup$ May 12, 2019 at 21:42

1 Answer 1


If you continually bet $10\%$ of capital with $0.1$ chance of returning $20x$ you will win almost surely. Your position after $n$ wins and $m$ losses only depends on $n$ and $m$, not on the order of wins and losses. Each win triples your bankroll, while each loss multiplies it by $0.9$. You will have $3^n\cdot 0.9^m$ times your original stake. The law of large numbers is on your side. After one win and nine losses your capital has multiplied by $3 \cdot 0.9^9\approx 1.16$. You will never run out of money because you never bet your whole bankroll. After a large number of games the winning fraction will be close to $10\%$ and you will be way ahead. The bet is in your favor-take it!

You might be interested to look up the Kelly criterion for how much to bet in a favorable case. If every dollar is equally useful to you, the highest expectation is to bet all your money each time. If twice as much money is not twice as good, you don't necessarily want to bet all your money.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .